我创建了一个由自定义适配器处理的自定义列表视图。此列表中的每个项目都显示图像和共享按钮。使用离子从外部源加载图像。这很好。
现在我想在用户点击按钮时分享图片。虽然我能够共享文本等,但即使在实现此代码后,我也无法共享这些外部图像:Sharing Remote Images
两件事:
if (drawable instanceof BitmapDrawable)
是假的,但我不知道为什么或如何解决这个问题非常感谢任何帮助或建议!
这是我的适配器的代码:
@Override
public View getView(int position, View convertView, ViewGroup parent) {
ViewHolder holder;
if(convertView == null) {
//brand new
convertView = LayoutInflater.from(mContext).inflate(R.layout.list_item, null);
holder = new ViewHolder();
holder.contentImageView = (ImageView) convertView.findViewById(R.id.contentImageView);
holder.contentLabel = (TextView) convertView.findViewById(R.id.contentLabel);
holder.contentShareButton = (Button) convertView.findViewById(R.id.contentShareButton);
convertView.setTag(holder);
}
else {
holder = (ViewHolder) convertView.getTag();
}
Content content = mContents[position];
Ion.with(mContext)
.load(content.getSrc()) //the external image url
.intoImageView(holder.contentImageView);
holder.contentLabel.setText(content.getTitle());
holder.contentShareButton.setTag(holder.contentImageView);
holder.contentShareButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
ImageView ivImage = (ImageView) v.getTag();
//ImageView ivImage = (ImageView) findViewById(R.id.ivResult);
// Get access to the URI for the bitmap
Uri bmpUri = getLocalBitmapUri(ivImage);
if (bmpUri != null) {
// Construct a ShareIntent with link to image
Intent shareIntent = new Intent();
shareIntent.setAction(Intent.ACTION_SEND);
shareIntent.putExtra(Intent.EXTRA_STREAM, bmpUri);
shareIntent.setType("image/*");
// Launch sharing dialog for image
mContext.startActivity(Intent.createChooser(shareIntent, "Share Image"));
} else {
Log.i("test", "Sharing failed, handler error.");
}
}
});
return convertView;
}
// Returns the URI path to the Bitmap displayed in specified ImageView
public Uri getLocalBitmapUri(ImageView imageView) {
// Extract Bitmap from ImageView drawable
Drawable drawable = imageView.getDrawable();
Bitmap bmp = null;
if (drawable instanceof BitmapDrawable){
bmp = ((BitmapDrawable) imageView.getDrawable()).getBitmap();
} else {
Log.i("test", "is null");
return null;
}
// Store image to default external storage directory
Uri bmpUri = null;
try {
File file = new File(Environment.getExternalStoragePublicDirectory(
Environment.DIRECTORY_DOWNLOADS), "share_image_" + System.currentTimeMillis() + ".png");
file.getParentFile().mkdirs();
FileOutputStream out = new FileOutputStream(file);
bmp.compress(Bitmap.CompressFormat.PNG, 90, out);
out.close();
bmpUri = Uri.fromFile(file);
} catch (IOException e) {
e.printStackTrace();
}
return bmpUri;
}
答案 0 :(得分:0)
将这四行添加到getLocalBitmapUri
- 函数让它对我有用:
bmp = Bitmap.createBitmap(drawable.getIntrinsicWidth(), drawable.getIntrinsicHeight(), Bitmap.Config.ARGB_8888);
Canvas canvas = new Canvas(bmp);
drawable.setBounds(0, 0, canvas.getWidth(), canvas.getHeight());
drawable.draw(canvas);
以下是包含更新代码的完整函数:
// Returns the URI path to the Bitmap displayed in specified ImageView
public Uri getLocalBitmapUri(ImageView imageView) {
// Extract Bitmap from ImageView drawable
Drawable drawable = imageView.getDrawable();
Bitmap bmp = null;
if (drawable instanceof BitmapDrawable){
bmp = ((BitmapDrawable) imageView.getDrawable()).getBitmap();
} else {
bmp = Bitmap.createBitmap(drawable.getIntrinsicWidth(), drawable.getIntrinsicHeight(), Bitmap.Config.ARGB_8888);
Canvas canvas = new Canvas(bmp);
drawable.setBounds(0, 0, canvas.getWidth(), canvas.getHeight());
drawable.draw(canvas);
}
// Store image to default external storage directory
Uri bmpUri = null;
try {
File file = new File(Environment.getExternalStoragePublicDirectory(
Environment.DIRECTORY_DOWNLOADS), "share_image_" + System.currentTimeMillis() + ".png");
file.getParentFile().mkdirs();
FileOutputStream out = new FileOutputStream(file);
bmp.compress(Bitmap.CompressFormat.PNG, 90, out);
out.close();
bmpUri = Uri.fromFile(file);
} catch (IOException e) {
e.printStackTrace();
}
return bmpUri;
}